Difference between revisions of "2006 AMC 12B Problems/Problem 2"

(Solution)
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<math>4\spadesuit 5=-9</math>
 
<math>4\spadesuit 5=-9</math>
  
<math>3\spadesuit -9=-72 \Rightarrow \text{(B)}</math>
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<math>3\spadesuit -9=-72 \Rightarrow \text{(A)}</math>
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=1|num-a=3}}
 
{{AMC12 box|year=2006|ab=B|num-b=1|num-a=3}}

Revision as of 23:12, 9 February 2008

Problem

For real numbers $x$ and $y$, define $x\spadesuit y = (x + y)(x - y)$. What is $3\spadesuit(4\spadesuit 5)$?

$\text {(A) } - 72 \qquad \text {(B) } - 27 \qquad \text {(C) } - 24 \qquad \text {(D) } 24 \qquad \text {(E) } 72$

Solution

$4\spadesuit 5=-9$

$3\spadesuit -9=-72 \Rightarrow \text{(A)}$

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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